Prove that $\operatorname{int}(A)$, $\operatorname{int}(X\backslash A)$ and $\partial(A)$ are pairwise disjoint sets whose union is $X$.

Question

I have been trying to prove this problem, but I am worried that my assumptions may not all be correct.

I would really appreciate it if you could offer some insight or possibly point out any assumptions that I made that are not correct.

Notations:

$\operatorname{int}(A)$ is used for the set of all interior points of $A$, and $\partial(A)$ are the boundary points of $A$

Question

Let $A$ be a subset of a metric space $X$.

Prove that: $\operatorname{int}(A)$, $\operatorname{int}(X\backslash A)$ and $\partial(A)$ are pairwise disjoint sets whose union is $X$.

My attempted proof

$A \cap X\backslash A = \emptyset$

Thus as $\operatorname{int}(A) \subseteq A$ and $\operatorname{int}(X\backslash A) \subseteq X\backslash A$, we have $\operatorname{int}(A) \cap \operatorname{int}(X\backslash A)= \emptyset$

This shows us $\operatorname{int}(A)$ and $\operatorname{int}(X\backslash A)$ are disjointed.

Now $\partial(A)= \overline{A} \cap \overline{X\backslash A}$

i)

$\operatorname{int}(A)\cap \partial(A)= \operatorname{int}(A)\cap\overline{A} \cap \overline{X\backslash A} = \operatorname{int}(A) \cap \overline{X\backslash A}$

Let $x\in \overline{X\backslash A}$, thus every nbd$(x)$ contains a point of $X\backslash A$

Suppose $x \in \operatorname{int}(A)$, but this will mean for any open ball $N$ of $x$, $\ N \cap X\backslash A \neq \emptyset$

Thus $N \nsubseteq A$ making $x \notin \operatorname{int}(A)$ but this is a contradiction thus $\operatorname{int}(A)\cap \partial(A) = \emptyset$

ii)

$int(X\backslash A)\cap bd(A)= int(X\backslash A)\cap\overline{A} \cap \overline{X\backslash A} = int(X\backslash A) \cap \overline{A}$

Let $a \in \overline{A}$ thus every nbd$(a)$ contains a point of $A$

Suppose $a \in \operatorname{int}(X\backslash A)$, but this will mean for any open ball $N$ of $a$, $\ N \cap A \neq \emptyset$

Thus $N \nsubseteq X\backslash A$ making $a \notin \operatorname{int}(X\backslash A)$ but this is a contradiction thus $\operatorname{int}(X\backslash A)\cap \partial(A) = \emptyset$

This shows that $(\operatorname{int}(A)\cup \operatorname{int}(X\backslash A))\cap \partial(A) = \emptyset$

Thus $\operatorname{int}(A)$, $\operatorname{int}(X\backslash A)$ and $\partial(A)$ are pairwise disjoint sets

This concludes my attempted proof

Answer 1

"Let $x\in \overline{X\backslash A}$, thus every nbd$(x)$ contains a point of $X\backslash A$

Suppose $x \in int(A)$, but this will mean for any open ball $N$ of $x$, $\ N \cap X\backslash A \neq \emptyset$"

It's =, not $\neq$ here: $\ N \cap X\backslash A$ = $\emptyset.$

Which would be a contradiction.

And the same holds for your second mentioned case ii)